Showing that $d\theta(\partial\Psi/\partial\theta)=1$ In Bachman's A Geometric Approach to Differential Forms, second edition, the following is exercise 8.4.

Often we parametrize surfaces in $\mathbb{R}^3$ by starting with cylindrical or spherical coordinates and expressing one of the parameters in terms of the other two. So, for example, by starting with spherical cooridnates and expressing $\rho$ as a function of $\theta$ and $\phi$, we may end up with a parametrisation $\Psi(\theta,\phi)=(x,y,z)$. However, we may also view $\theta(x,y,z)$ as a function on $\mathbb R^3$ that gives the $\theta$ coordinate of the point $(x,y,z)$, when it is expressed in spherical coordinates. Hence, it is often the case that we have parametrisations of the form $\Psi(\theta,\phi)$ such that $$\theta(\Psi(\theta,\phi))=\theta\text{ and }\phi(\Psi(\theta,\phi))=\phi.$$When we think of $\theta(x,y,z)$ as a 0-form on $\mathbb R^3$, we can differentiate to get a 1-form $d\theta$. Show that $$d\theta\left(\frac{\partial\Psi}{\partial\theta}\right)=d\phi\left(\frac{\partial\Psi}{\partial\phi}\right)=1$$and$$d\theta\left(\frac{\partial\Psi}{\partial\phi}\right)=d\phi\left(\frac{\partial\Psi}{\partial\theta}\right)=0.$$

I think I am mainly confused by how $d\theta$ acts on $\partial\Psi/\partial\theta$ (and similarly with $\phi$ replacing $\theta$). If given an explicit 1-form on some tangent space $T_pX$, say $$\omega_p=a_1dx+a_2dy+a_3dz,$$ and explicitly knowing some tangent vector $v=\langle b_1,b_2,b_3\rangle_p$ in that tangent space, I know how to compute $$\omega_p(v)=a_1b_1+a_2b_2+a_3b_3.$$
However here I am not sure how to geometrically visualise $\partial\Psi/\partial\theta$ as a tangent vector to $\operatorname{im}\Psi$ (the surface being paramatrised by $\Psi$), how to compute it, and hence how to find $d\theta(\partial\Psi/\partial\theta)$. For example since $\theta=\arctan(y/x)$ then I find that $$d\theta=-\frac{y}{x^2+y^2}dx+\frac{x}{x^2+y^2}dy$$ and if we let $\Psi=(\Psi_1,\Psi_2,\Psi_3)$ I can find that $$\frac{\partial\Psi}{\partial\theta}=\left\langle\frac{\partial\Psi_1}{\partial\theta},\frac{\partial\Psi_2}{\partial\theta},\frac{\partial\Psi_3}{\partial\theta}\right\rangle.$$
Now, I evaluate that
$$d\theta\left(\frac{\partial\Psi}{\partial\theta}\right)=-\frac{y}{x^2+y^2}\frac{\partial\Psi_1}{\partial\theta}+\frac{x}{x^2+y^2}\frac{\partial\Psi_2}{\partial\theta}$$
which is clearly not $1$. Thus I get a feeling that there is some huge conceptual gap which is preventing me from solving the problem, one which I cannot seem to find myself.
Thanks in advance for any help---just pointing out some concept or idea that I'm missing, even if not a full solution, would already be of great assistance!
 A: Given $f : \mathbb{R}^2 \to \mathbb{R}$, we have $\frac{\partial f}{\partial x}(x,y) = d_{(x,y)} f \begin{pmatrix} 1 \\ 0\end{pmatrix}$. Therefore we have the following,
$$ \frac{\partial}{\partial \theta} \left( \theta \circ \Psi(\theta, \phi) \right) = d(\theta \circ \Psi)(\theta, \phi) \begin{pmatrix} 1 \\ 0 \end{pmatrix}$$
Now use chain rule on the RHS,
$$\begin{align*} d(\theta \circ \Psi)(\theta, \phi) \begin{pmatrix} 1 \\ 0 \end{pmatrix} &= d_{\Psi(\theta, \phi)} \theta \circ d_{(\theta, \phi)} \Psi \begin{pmatrix} 1 \\ 0 \end{pmatrix} \\ & = d\theta \begin{pmatrix} \frac{\partial \Psi}{\partial \theta} & \frac{\partial \Psi}{\partial \phi} \end{pmatrix}\begin{pmatrix} 1 \\ 0 \end{pmatrix} & \\ = d \theta \left(\frac{\partial \Psi}{\partial \theta}\right) \end{align*}$$
But isn't this $1$ since $\frac{\partial}{\partial \theta} \left( \theta \circ \Psi(\theta, \phi) \right) = \frac{\partial}{\partial \theta} (\theta)$ ? Take the other partial to get the zero result. You can then repeat this with $\phi$.
